The bird at right watches the waves. If the portion of a wave between two crests passes the pole each second. Wavelength is 1m.

The temperature of air changes from 0 to 10°C while its velocity changes from zero to a final velocity, and its elevation change

aliya0001 [1]

Answer:

Final velocity = 119.83 m/s

Final elevation = 731.9 m

Explanation:

We are told that temperature of air changes from 0 to 10°C

Thus;

Change in temperature; ΔT = 10 - 0 = 10°C

Also, its velocity changes from zero to a final velocity. Thus;

v1 = 0 m/s

v2 is unknown

Also, its elevation changes from zero to a final elevation.

So, z1 = 0 and z2 is unknown

Now, we want to find v2 and z2 when the internal, kinetic and potential energy are equal.

Thus Equating the formula for both kinetic and internal energy gives;

½m(v2² - v1²) = mc_v•ΔT

m will cancel out and v1 is zero to give;

v2² = 2c_v•ΔT

v2 = √(2c_v•ΔT)

Where c_v is specific heat of constant volume of air with a constant value of 718 J/Kg.K

Thus;

v2 = √(2 × 718 × 10)

v2 = √14360

v2 = 119.83 m/s

To find z2, we will equate potential energy formula to that of the internal energy.

Thus;

mg(z2 - z1) = mc_v•ΔT

m will cancel out and since z1 is zero, then we have;

z2 = (c_v•ΔT)/g

z2 = 718 × 10/9.81

z2 = 731.9 m

4 0

3 years ago

Can someone please help, ty!

lyudmila [28]

Answer:hypothesis: the plant died of lack of light,moisture,or water

Explanation:

to test my hypothesis i put a plant in a room at room temp and repeat how i grew it,and observe what went wrong. hope this helps=)

5 0

4 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes:

$v_2(32.6\,\mathrm s)=-196\frac{\rm m}{\rm s}$

That is, the rocket has a velocity of 196 m/s in the downward direction as it hits the ground.

3 0

3 years ago

A car slows from 27 m/s to 5 m/s with a constant acceleration for 6.87 s. What is the car’s acceleration?

Kobotan [32]

In this case, the movement is uniformly delayed (the final rapidity is less than the initial rapidity), therefore, the value of the acceleration will be negative.

1. The following equation is used:

a = (Vf-Vo)/ t

a: acceleration (m/s2)

Vf: final rapidity (m/s)

Vo: initial rapidity (m/s)

t: time (s)

2. Substituting the values in the equation:

a = (5 m/s- 27 m/s)/6.87 s

3. The car's acceleration is:

a= -3.20 m/ s^2

5 0

3 years ago

PLEASE HELP ME!

PilotLPTM [1.2K]

Answer:

a) A = 3 cm, b) T = 0.4 s, f = 2.5 Hz,

2) A standing wave the displacement of the wave is canceled and only one oscillation remains

Explanation:

a) in an oscillatory movement the amplitude is the highest value of the signal in this case

A = 3 cm

b) the period of oscillation is the time it takes for the wave to repeat itself in this case

T = 0.4 s

the period is the inverse of the frequency

f = 1 /T

f = 1 /, 0.4

f = 2.5 Hz

2) a traveling wave is a wave for which as time increases the displacement increases, in the case of a transverse wave the oscillation is perpendicular to the displacement and in the case of a longitudinal wave the oscillation is in the same direction of the displacement.

A standing wave occurs when a traveling wave bounces off some object and there are two waves, one that travels in one direction and the other that travels in the opposite direction. In this case, the displacement of the wave is canceled and only one oscillation remains.

8 0

3 years ago